I am very new to Lie groups and manifolds. In my self study, many times I have come across "differential (or derivative) of the left translation" (for example, here). I don't fully understand what is meant by differential/derivative here. I would appreciate any attempt to explain this concept both formally and intuitively.

For example, what does the differential/derivative here mean and how is it defined? Does the definition require charts in general (I have seen how smooth maps are defined between two manifolds)? Would it have a more convenient form in the special case of matrix Lie groups? What does it do?

Part of the answer given to this question is definitely relevant. But I am not sure where does $$dL_{g}(v) = \frac{d}{dt}\bigg|_{t=0} L_{g}\exp(tv)$$ come from? (I know that $\exp(tv)$ is a curve on $G$ that passes through the identity element at $t=0$ and $v$ is its tangent vector at the identity). Is this the definition of differential of $L_{g}$? Is there a backstory?

To make this question more concrete, let me set up a standard notation. Let $G$ be a matrix Lie group. $L_g : G \to G$ for any $g \in G$ maps an element of $G \ni p$ to $ L_g(p) := g p$. Then, the differential of $L_g$, denoted by $dL_g$ is a mapping from a tangent space at any $p \in G$ to the tangent space at $L_g(p)$.

  • $\begingroup$ $dL_g$ is the pushforward of the map $L_g$. How is that defined? $\endgroup$ Commented Sep 30, 2019 at 7:01
  • $\begingroup$ Just to be clear, you are familiar with treating $dL_g$ as a map between tangent spaces, but you want clarification on how it can be written in terms of the exponential map? The formula you give only gives an expression for $dL_g$ as a map from $T_eG$ to $T_gG$. $\endgroup$
    – Kajelad
    Commented Sep 30, 2019 at 19:23
  • $\begingroup$ Technically, tangent vectors are derivations. In this sense, the above definition is incomplete. $\endgroup$ Commented Sep 30, 2019 at 20:11
  • $\begingroup$ Thanks for the help folks. @Kajelad I wasn't familiar with pushforward - it makes sense. To make this complete, do I need to mention a point on $G$ (say, $p$) at which $d L_g$ is being calculated? In that first I need the curve that goes through $p$ at $t = 0$ and has a tangent vector $v$. In this case, $p \exp(t p^{-1} v)$ would be such a curve. Mapping this curve via $L_g$ gives $g p \exp(t p^{-1} v)$. The tangent vector at $ t = 0 $ is then going to be $ g v$. And this holds for any $p$. Is this correct? $\endgroup$
    – newbie777
    Commented Oct 2, 2019 at 4:27
  • $\begingroup$ @OliverJones Thanks for the help! I haven't fully studied the alternative definition of tangent vectors as derivatives. Are these two definitions equivalent and equally "formal"? Perhaps related to this question: what is allowing me here to take the "usual" derivative $ \frac{d}{d t} $ from a curve such as $t \mapsto \exp(t v) \in G$ without involving charts? I guess this has something to do with the fact that $ G $ is assumed to be a matrix Lie group - can you please elaborate? Thanks! $\endgroup$
    – newbie777
    Commented Oct 2, 2019 at 4:36

1 Answer 1


As you note, $dL_g:T_hG\to T_{gh}G$ (or the differential of any smooth map) is already a well-defined linear a map between tangent spaces without further reference to the group structure. In terms of velocities of curves, it is defined by: $$ dL_g\left(\left.\frac{d}{dt}\gamma(t)\right|_{t=t_0}\right)=\left.\frac{d}{dt}\left(L_g(\gamma(t))\right)\right|_{t=t_0} $$ Equivalently, in terms of derivations: $$ [dL_g(V_h)](f)=V_h(f\circ L_g) $$ These two definitions of tangent vectors are equivalent: we may equate every velocity with a derivation given by $$ \left(\left.\frac{d}{dt}\gamma(t)\right|_{t=t_0}\right)(f)=\left.\frac{d}{dt}\left(f(\gamma(t))\right)\right|_{t=t_0} $$ If this isn't already familiar, it might be worth checking that the above definitions of the differential agree.

Understanding its relationship with the exponential map will require establishing a few facts about the Lie algebras and the exponential map.

The Lie algebra $\text{Lie}(G)$, can be identified with two vector spaces:

  • The tangent space $T_eG$
  • The space of left-invarant vector fields on $G$

To go from the first to the second, we can just evaluate the vector field at $e$. To go from a tangent vector $v\in T_e M$ to a vector field $V$, we let $V_e=v$, and the value of $V$ at any other point $g$ must be $V_g=dL_gv$ by left invariance. Because of this, we often use the same symbol to refer to both a tangent vectors at $e$ and the corresponding left-invariant vector field. Instead, I'll treat elements $V\in\text{Lie}(G)$ as left-invariant vector fields and use subscripts $V_e\in T_eM$ to refer to their values at particular points.

The exponential map $\exp:\text{Lie}(G)\to G$ is given by $$ \exp(V)=\gamma_V(1) $$ Where $\gamma_V$ is the unique integral curve of the left-invariant vector field $V$ satisfying $\gamma_V(0)=e$ and $\frac{d}{dt}\gamma(t)=V_{\gamma(t)}$ (often called the one parameter subgroup generated by $V$). A consequence of this is that $\gamma_V(t)=\exp(tV)$.

Let $U_e\in T_e G$, and $U$ be the corresponding left-invariant vector field. We can use definition of the differential, and the fact that $\left.\frac{d}{dt}\exp(tU)\right|_{t=0}=U_e$ to arrive at the formula you give: $$ \left.\frac{d}{dt}\left[L_g\exp(tU)\right]\right|_{t=0}=dL_g\left.\frac{d}{dt}\exp(tU)\right|_{t=0}=dL_gU_e $$ We see, somewhat more explicitly, this is only an expression for how $dL_g$ acts on tangent vectors at the identity.

  • $\begingroup$ Thanks @Kajelad - and can you please help me to understand why are we allowed to talk about $\frac{d}{dt}$ of a curve that lives on the matrix Lie group? (as far as I know, $\frac{d}{dt}$ would be meaningless for curves on abstract manifolds - I'm trying to understand formally what property of matrix Lie groups allows us to do this? is it the fact that they are embedded in a Euclidean space? can we do the same thing for sphere?) $\endgroup$
    – newbie777
    Commented Oct 5, 2019 at 19:46
  • $\begingroup$ The velocity of a curve $\frac{d}{dt}\gamma$ is a tangent vector; there are several definitions which work on arbitrary manifolds. The most common one is that tangent vectors are (pointwise) derivations (i.e. pointwise directional derivatives): maps from smooth real-valued functions to real numbers satisfying $$v_p(fg)=f(p)v_p(g)+g(p)v_p(f)$$ where $f,g$ are smooth real valued functions, $p$ is a point on the manifold, and $v_p$ is a derivation at $p$. We can think of the velocity of a curve at a point as a derivation, as in the answer. $\endgroup$
    – Kajelad
    Commented Oct 5, 2019 at 20:45
  • $\begingroup$ Thanks again @Kajelad. I understand there is a more general/formal definition of tangent vectors as a derivation; i.e., in general $\frac{d}{dt} \gamma(t)$ is not even meaningful according to the definition of $\frac{d}{dt}$. But in the case of matrix Lie groups it seems that we can pretend $\gamma(t)$ maps to $\mathbb{R}^{n^2}$ and simply take the usual derivative? i.e., if $\gamma(t) \in \mathrm{SO}_n$ is a $n \times n$ matrix-valued function, I can easily compute tangent vectors to $\gamma(t)$ by explicitly computing the usual derivative of the $n \times n$ matrix-valued function wrt $t$. $\endgroup$
    – newbie777
    Commented Oct 5, 2019 at 21:10
  • 1
    $\begingroup$ Yes. What you're effectively doing there is using the matrix elements as a set of global coordinates, with the corresponding partial derivatives as a bases for the tangent spaces. The matrix valued function $\frac{d}{dt}\gamma(t)$ is the coordinate form of the abstract tangent vector $\frac{d}{dt}\gamma$, where each element of the matrix is the component in the corresponding partial derivative. $\endgroup$
    – Kajelad
    Commented Oct 5, 2019 at 21:48

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